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Lecture 21 - Feb 27

# Lecture 21 - Feb 27 - Friday February 27 Lecture 21 Finding...

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Friday February 27 Lecture 21 : Finding bases: Expanding a linearly independent set to a basis of a vector space. (Refers to section 4.4) Expectations: 1. Apply an algorithm to expand a linearly independent set of vectors to a basis of V . 21.1 Proposition Let U = { v 1 , v 2 , .... , v k } be a linearly independent set of vectors in the vector space V such that span( v 1 , v 2 , .... , v k ) is NOT all of V . Then, for any vector v in V / span( U ), the set { v 1 , v 2 , .... , v k , v } forms a linearly independent set. Proof (Outline): o Let U = { v 1 , v 2 , .... , v k } be linearly independent and v be a vector not in the span of U . o Suppose α 1 v 1 + α 2 v 2 +..., + α k v k + α v = 0 . o We claim α = 0: . Suppose α ≠ 0. Then v = ( α 1 / α ) v 1 + ( α 2 / α ) v 2 + , ..., + ( α k / α ) v k span( U ), a contradiction. Thus α = 0. o Also α 1 v 1 + α 2 v 2 +..., + α k v k = 0 implies α i = 0 for all i. Why? o Thus α i = 0 for all i and α = 0 . o Thus { v 1 , v 2 , .... , v k , v } forms a linearly independent set. ° 21.2.Consequence of Proposition 21.1 For any finite dimensional vector space V , a linearly independent set of vectors can be completed to a basis for V .

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Lecture 21 - Feb 27 - Friday February 27 Lecture 21 Finding...

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